Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 18 July 2014

Lecture 23

A field is a set $𝔽$ with operations $$\begin{array}{cccc}+:& 𝔽\times 𝔽& ⟶& 𝔽\\ & (a,b)& ⟼& a+b\end{array}\begin{array}{ccc}𝔽\times 𝔽& ⟶& 𝔽\\ (a,b)& ⟼& ab\end{array}$$ such that

(a)	if $a,b,c\in 𝔽$ then $(a+b)+c=a+(b+c),$
(b)	if $a,b\in 𝔽$ then $a+b=b+a,$
(c)	there exists $0\in 𝔽$ such that if $a\in 𝔽$ then $0+a=a$ and $a+0=a,$
(d)	if $a\in 𝔽$ then there exists $-a\in 𝔽$ such that $a+(-a)=0$ and $(-a)+a=0,$
(e)	if $a,b,c\in 𝔽$ then $\left(ab\right)c=a\left(bc\right),$
(f)	if $a,b,c\in 𝔽$ then $$(a+b)c=ac+bc\text{and}c(a+b)=ca+cb,$$
(g)	there exists $1\in 𝔽$ such that if $a\in 𝔽$ then $1·a=a$ and $a·1=a,$
(h)	if $a\in 𝔽$ and $a\ne 0$ then there exists $a^{-1}\in 𝔽$ such that $a{a}^{-1}=1$ and $a^{-1}a=1,$
(i)	if $a,b\in 𝔽$ then $ab=ba\text{.}$

The positive integers is the set $${\mathbb{Z}}_{>0}=\{1,1+1,1+1+1,1+1+1+1,\dots \}$$ with addition given by concatenation so that $$(1+1)+(1+1+1)=1+1+1+1+1,\text{for example.}$$ The counting by $k$ function is $m_k:{\mathbb{Z}}_{>0}\to {\mathbb{Z}}_{>0}$ given by $$m_k\left(1\right)=k\text{and}{m}_k(a+b)=m_k\left(a\right)+m_k\left(b\right)\text{.}$$ For example: $$\begin{array}{cccc}{m}_5:& {\mathbb{Z}}_{>0}& ⟶& {\mathbb{Z}}_{>0}\\ & 1& ⟼& 5\\ & 2& ⟼& 10\\ & 3& ⟼& 15\\ & ⋮& ⟼& ⋮\end{array}$$ Multiplication on ${\mathbb{Z}}_{>0}$ is given by $$\begin{array}{ccc}{\mathbb{Z}}_{>0}\times {\mathbb{Z}}_{>0}& ⟶& {\mathbb{Z}}_{>0}\\ (k,\ell )& ⟼& m_k\left(\ell \right)\end{array}$$ The nonnegative integers is the set $${\mathbb{Z}}_{\ge 0}=\{0,1,2,3,\dots \}$$ where $2$ really means $1+1,$ $3$ really means $1+1+1,$ etc.

The integers is the set $$\mathbb{Z}=\{\dots ,-3,-2,-1,0,1,2,\dots \}$$ with addition and multiplication determined from ${\mathbb{Z}}_{>0}$ and the axioms of a field, except $\left(h\right)\text{.}$

Define a relation $\le$ on $\mathbb{Z}$ by $x\le y$ if there exists $n\in {\mathbb{Z}}_{\ge 0}$ such that $x+n=y\text{.}$

The rational numbers is the set $$\mathbb{Q}=\left\{\frac{a}{b}\hspace{0.17em}\right|\hspace{0.17em}a,b\in \mathbb{Z},b\ne 0\}$$ with $$\frac{a}{b}=\frac{c}{d}\text{if}ad=bc,$$ and $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\text{and}\frac{a}{b}·\frac{c}{d}=\frac{ac}{bd}\text{.}$$

The real numbers $ℝ$ is the set $$ℝ={ℝ}_{\le 0}\cup {ℝ}_{\ge 0},$$ where $$\begin{array}{ccc}{\displaystyle {ℝ}_{\ge 0}}& =& {\displaystyle \{a_{\ell }{a}_{\ell -1}\cdots a_1{a}_0·a_{-1}{a}_{-2}\text{.}\hspace{0.17em}|\hspace{0.17em}\ell \in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}}\\ {\displaystyle {ℝ}_{\le 0}}& =& {\displaystyle \{-a_{\ell }{a}_{\ell -1}\cdots a_1{a}_0\text{.}{a}_{-1}{a}_{-2}\cdots \hspace{0.17em}|\hspace{0.17em}\ell \in {\mathbb{Z}}_{\ge 0},a_i\in \{0,1,\dots ,9\}\}}\end{array}$$ and $$a=b\text{if}a-b=0.000\dots$$ where addition and multiplication are given by $a+b$ to an accuracy of $k$ decimal places is $a_{\ge (-k-1)}+b_{\ge (-k-1)}$ (addition in $\mathbb{Q}\text{),}$ with $$a_{\ge -k}=a_{\ell }{a}_{\ell -1}\cdots a_1{a}_0\text{.}{a}_{-1}{a}_{-2}\cdots a_{-k}000\dots$$ and $ab$ to an accuracy of $k$ decimal places is $a_{\ge (-m-k)}·b_{\ge (-\ell -k)}$ (multiplication in $\mathbb{Q}\text{)}$ if ?????????

The complex numbers $ℂ$ is the set $$ℂ=\{a+bi\hspace{0.17em}|\hspace{0.17em}a,b\in ℝ\}\text{with}{i}^2=-1,$$ the addition and multiplication in $ℝ$ and the properties of a field determining the addition and multiplication in $ℂ\text{.}$

For example: $$(3+2i)+(5+7i)=8+9i$$ and $$\begin{array}{ccc}{\displaystyle (3+2i)(5+7i)}& =& {\displaystyle 15+21i+10i+14i^2}\\ & =& {\displaystyle (15-14)+31i=1+31i}\end{array}$$ and $${(3+2i)}^{-1}=\frac{1}{(3+2i)}\frac{(3-2i)}{(3-2i)}=\frac{1}{9-4}(3-2i)=\frac{3}{5}-\frac{2}{5}i\text{.}$$ Graphing: $$\begin{array}{c} 3-2i 2-3i i ℝ \end{array}$$

Let $a=210.98765432109876543210987\dots$ and $b=b=100101.101001000100001\dots \text{.}$ Compute $a+b$ and $a·b$ to $10$ decimal places. Well $$\frac{\hfill {\displaystyle \genfrac{}{}{0ex}{}{\hfill \mathrm{210.987654321098765...}}{\hfill \mathrm{100101.101001000100001...}}}}{\hfill {\displaystyle \genfrac{}{}{0ex}{}{\hfill 100312.088655321198766\phantom{\mathrm{...}}}{\hfill \stackrel{\text{or}}{7}\phantom{\mathrm{...}}}}}$$ and $$b={10}^5+{10}^2+1+{10}^{-1}+{10}^{-3}+{10}^{-6}+{10}^{-10}+{10}^{-15}+{10}^{-21}+\cdots \text{.}$$ So $$\begin{array}{ccc}{\displaystyle {10}^{5}a}& =& {\displaystyle \mathrm{21098765.432109876543210...}}\\ {\displaystyle {10}^{2}a}& =& {\displaystyle \mathrm{21098.765432109876543...}}\\ \multicolumn{3}{c}{\underset{}{─}}\\ {\displaystyle ({10}^5+{10}^2)a}& =& {\displaystyle \mathrm{21119864.197541986419753...}}\\ {\displaystyle a}& =& {\displaystyle \mathrm{210.987654321098765...}}\\ \multicolumn{3}{c}{\underset{}{─}}\\ {\displaystyle ({10}^5+{10}^2+1)a}& =& {\displaystyle \mathrm{21120075.185195307518518...}}\\ {\displaystyle {10}^{-1}a}& =& {\displaystyle \mathrm{21.098765432109876...}}\\ \multicolumn{3}{c}{\underset{}{─}}\\ & & {\displaystyle 21120096.283960739628394\phantom{\mathrm{...}}}\end{array}$$ etc.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100503Lect23.pdf and was given on 3 May 2010.

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